By symmetry, the centre of your sphere lies on the "bi-altitude" line joining the mid-points of near and far tetrahedron edges: choose this line as our vertical. All four planes lie at the same angle to the it, therefore their points of tangency to the sphere all lie on the same horizontal plane. By symmetry these points are the vertices of a plane rhombus; and since they lie on the sphere, the rhombus is cyclic. Therefore it is a square; the near and far edges of the tetrahedron meet on the bi-altitude; and your regular tetrahedron degenerates to a single point. However, I now realise that here is a sense in which the `sphere' in this case has merely retreated to (inversive) infinity, making it a most unsuitable candidate for generalisation! Should perhaps have read (B) That m = 2 is not always possible when n = 3 (for a tetrahedron) (now that it's been comprehensively cooked and spoiled); (C) What happens for m = 3 and n = 4 (for a pentatope); WFL On 4/7/16, Allan Wechsler <acwacw@gmail.com> wrote:
Forgive me for still not seeing this; it still looks feasible to me in the regular case. Suppose I put a tiny marble exactly halfway along an edge, exteriorly tangent to the two faces incident to that edge. Now, keeping the center of the sphere on the perpendicular bisector plane of that edge, and keeping it tangent to those two planes, let it inflate slowly. It has to move away from the tetrahedron in order to maintain tangency. At some point in this process, surely the sphere will kiss the two planes that form the ends of the trough. What am I missing? If this goes on I will have to calculate coordinates!
On Wed, Apr 6, 2016 at 6:00 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Your picture is surely partly correct, in that any planes may simply be adjusted so that they meet the given sphere and still form a trough outside some tetrahedron.
However given a regular tetrahedron, that trough turns out to be the wrong shape to accommodate any sphere!
I initially thought I had a rather slick argument supporting (B), but further contemplation has revealed the (inevitable) flaw.
So (B) is probably false in general, and a little more spadework is indicated --- customary apologies!
WFL
On 4/6/16, Allan Wechsler <acwacw@gmail.com> wrote:
Does (B) have a typo? I have a pretty good visualization of a sphere tangent to the inside of two faces and the outside of two others. It hovers just outside the common edge of the two faces it's exteriorly tangent to, and the other two faces cut off the ends of the trough it rests in. I do see that m=3 is impossible for the tetrahedron.
On Wed, Apr 6, 2016 at 12:40 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I finally wrote up properly all the stuff I've been doing about inspheres and exspheres of a simplex. Finished. Done & dusted. Off my plate.
But as I sat down again to start writing up P*nc*l*t's perishin' p*r*sm, an obnoxious inspiration intruded. Why shouldn't a simplex in Euclidean space also sport an ` m-fold' exsphere, tangent to m facets on their exterior side, and to n-m+1 on their interior side, where perhaps m
1 ?
In particular, show
(A) That m = 2 is impossible when n = 2 (for a triangle);
(B) That m = 2 is impossible when n = 3 (for a tetrahedron);
(C) What happens for m = 2 and n = 4 (for a pentatope);
(D) What happens for general 0 <= m <= n+1 ?
Fred Lunnon
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